If `tan^(-1)x+tan^(-1)y=(pi)/(4)`, then `cot^(-1)x+cot^(-1)y=`

Options

A. `(pi)/(2)`
B. `(3pi)/(4)`
C. `pi`
D. `(5pi)/(4) `

Answer

To solve the problem, we need to find the value of \( \cot^{-1}x + \cot^{-1}y \) given that \( \tan^{-1}x + \tan^{-1}y = \frac{\pi}{4} \).

### Step-by-step Solution:

1. **Start with the given equation:**
\[ \tan^{-1}x + \tan^{-1}y = \frac{\pi}{4} \]

2. **Use the property of inverse tangent:**
We know that:
\[ \tan^{-1}x + \tan^{-1}y = \frac{\pi}{4} \implies \tan\left(\tan^{-1}x + \tan^{-1}y\right) = \tan\left(\frac{\pi}{4}\right) \]
Since \( \tan\left(\frac{\pi}{4}\right) = 1 \), we can write:
\[ \frac{x + y}{1 - xy} = 1 \]

3. **Cross-multiply to simplify:**
\[ x + y = 1 - xy \]

4. **Rearranging the equation:**
\[ xy + x + y + 1 = 2 \]
This can be rewritten as:
\[ xy + x + y + 1 - 2 = 0 \implies xy + x + y - 1 = 0 \]

5. **Now, we need to find \( \cot^{-1}x + \cot^{-1}y \):**
We use the identity:
\[ \cot^{-1}x + \cot^{-1}y = \frac{\pi}{2} - \left(\tan^{-1}x + \tan^{-1}y\right) \]

6. **Substituting the known value:**
Since \( \tan^{-1}x + \tan^{-1}y = \frac{\pi}{4} \):
\[ \cot^{-1}x + \cot^{-1}y = \frac{\pi}{2} - \frac{\pi}{4} \]

7. **Calculating the final value:**
\[ \cot^{-1}x + \cot^{-1}y = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \]

### Final Answer:
Thus, the value of \( \cot^{-1}x + \cot^{-1}y \) is:
\[ \frac{3\pi}{4} \]